destroyer ochem #33

[hokie4life]

why does the last choice only have one enantiomer? can't you flip the bromies like you did in the choice before to get more than one enantiomer since we have more than one chiral carbon?

doesn't more than one chiral carbon usually mean more than one enantiomer?

 

[userah]

which version of the book do you have? it doesn't seem to match up with the question i have. mine's the june edition

 

[cskpeter]

Bromination is anti and as a result it will have either (1S,2S) or (1R,2R) only.
No (R,S) or (S,R) -> No diastereomers

The choice (c) has an ethyl attached on the third carbon of the cyclopentane which acts as another chiral center.

 

[hokie4life]

Bromination is anti and as a result it will have either (1S,2S) or (1R,2R) only.
No (R,S) or (S,R) -> No diastereomers

The choice (c) has an ethyl attached on the third carbon of the cyclopentane which acts as another chiral center.

thank you for the response! i'm still not sure i understand though...

why can't we get R,S and S,R? is it because you can make meso compounds in choice D? bu then how do you get RR and SS? won't those be meso?

also-how did you know RR and SS were the enantiomers did you do it out? cuz for enantiomers isn't it just that they have to be opposites? like if one is RR the other is SS or if one is RS the other is SR? and then the diasteriomers are the ones that dont' match like RR and SR or SS and RS?

also, is there a general rule like if
-one chiral center = one enatiomer
-2 chiral centers = a pair of enatiomers, and you can get diasteriomers as long as you can't get meso cmpds?

 

[cskpeter]

Let me elaborate more on this.

First, when (D) cyclopentene is brominated, it will have two Bromines attached to carbon 1 and 2 of the cyclopentane. Those two Bromines are anti; if the first one is attached on dash and the other MUST be attached on wedge. Let's say the top carbon of the cyclopentane is the C1 on the solution's drawing. The configuration would be then (1S, 2S) according to the priority rule. You know that right? What if wedges and dashes are changed, i.e. the first bromine is on wedges and the other on dash? The configuration is now (1R, 2R). Now we have 2 stereoisomers for (D) and they are enantiomers only (1S, 2S) <--> (1R, 2R). But is (1S, 2R) or (1R, 2S) possible? No, because two bromines can be attached only in anti, not in syn. Don't be confused with meso and enantiomer concepts. See problem 37 for more clarification.

Second, don't forget that the enantiomers and diastereomers are relative terms. If we have (1R, 2R), its enantiomer is (1S, 2S) and its diastereomers are (1R, 2S) or (1S, 2R). But if we started with (1R, 2S), its enantiomer is (1S, 2R) and diastereomers are (1R, 2R) or (1S, 2S). I wouldn't go more about the definition.

Finally, the choice (C) is similar to (D) but with one difference. The ethyl group attached gives another chiral center to (C) when it's brominated to 1,2-dibromo-3-ethlycyclopentane. As discussed before, bromination is anti yielding either (R,R) or (S,S) and the third chiral center ethly group could be (R) or (S). For the simplicity's sake, think bromination has 2 possible choices and the ethyl group also has 2, which gives us 4 possible stereoisomer sets (2x2 =4). The possible stereoisomers are (1R, 2R, 3R), (1R, 2R, 3S), (1S, 2S, 3R), and (1S, 2S, 3S), all 4 of them. (1R, 2R, 3R) and (1S, 2S, 3S) are enantiomers. (1R, 2R, 3S) and (1S, 2S, 3R) are diastereomers to (1R, 2R, 3R) or (1S, 2S, 3S). So, choice (C) has diastereomers.

 

[hokie4life]

wow this is great! this makes SO much more sense THANK YOU SOOOO MUCH!!! i wasn't including that bromines add anti so that really threw it off. This is really great and clears it up well...thank you! 🙂

Let me elaborate more on this.

First, when (D) cyclopentene is brominated, it will have two Bromines attached to carbon 1 and 2 of the cyclopentane. Those two Bromines are anti; if the first one is attached on dash and the other MUST be attached on wedge. Let's say the top carbon of the cyclopentane is the C1 on the solution's drawing. The configuration would be then (1S, 2S) according to the priority rule. You know that right? What if wedges and dashes are changed, i.e. the first bromine is on wedges and the other on dash? The configuration is now (1R, 2R). Now we have 2 stereoisomers for (D) and they are enantiomers only (1S, 2S) <--> (1R, 2R). But is (1S, 2R) or (1R, 2S) possible? No, because two bromines can be attached only in anti, not in syn. Don't be confused with meso and enantiomer concepts. See problem 37 for more clarification.

Second, don't forget that the enantiomers and diastereomers are relative terms. If we have (1R, 2R), its enantiomer is (1S, 2S) and its diastereomers are (1R, 2S) or (1S, 2R). But if we started with (1R, 2S), its enantiomer is (1S, 2R) and diastereomers are (1R, 2R) or (1S, 2S). I wouldn't go more about the definition.

Finally, the choice (C) is similar to (D) but with one difference. The ethyl group attached gives another chiral center to (C) when it's brominated to 1,2-dibromo-3-ethlycyclopentane. As discussed before, bromination is anti yielding either (R,R) or (S,S) and the third chiral center ethly group could be (R) or (S). For the simplicity's sake, think bromination has 2 possible choices and the ethyl group also has 2, which gives us 4 possible stereoisomer sets (2x2 =4). The possible stereoisomers are (1R, 2R, 3R), (1R, 2R, 3S), (1S, 2S, 3R), and (1S, 2S, 3S), all 4 of them. (1R, 2R, 3R) and (1S, 2S, 3S) are enantiomers. (1R, 2R, 3S) and (1S, 2S, 3R) are diastereomers to (1R, 2R, 3R) or (1S, 2S, 3S). So, choice (C) has diastereomers.

 

[cskpeter]

Glad to hear that!😉

 

[queencrown]

awww..I made a search about question #33 and got this answer from 1 YEAR AGO!!! excellent answer!! 👍